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archeryguru2000
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Homework Statement
Medium lubricating oil, of specific gravity 0.860, is pumped through 300 m of horizontal 50-mm-diameter pipe at the rate of 0.00114 m3/s. If the drop in the pressure is 200 kPa, the absolute viscosity of the oil in N×s/m2 is...____________?
specific gravity (SG) = 0.86
length (l) = 300 m
diameter (d) = .05 m
flow rate (Q) = 0.00114 m^3/s
Pressure drop (P) = 200E3 Pa
Homework Equations
I know that:
shear stress (t) = viscosity (mu) * velocity gradient
where:
velocity gradient = (change in velocity) / (change in height)
The Attempt at a Solution
I'm not quite sure how to begin with this. I have a solution from my prof. (this is a question on a sample/practice exam for our final)... but I have NO clue where he's coming up with this equation:
P = 32 * (mu) & (l) * [(v)/(.05m)^2]
Where on Earth does the 32 come from? What significance does the (.05m)^2 have... without pi/4 anyway? Can anyone make any sense of this? The solution is supposed to be... approximately 0.09 (N s) / (m)^2. Let me know if anybody has any ideas.
Thanks,
~~Chad~~